Equation of a plane

A plane in three-dimensional space is determined by:

• One point on the plane and a normal direction, or
• Three non-collinear points, or
• Three intercepts on the coordinate axes (in normal form).

We develop all three forms.

Normal (vector) form

A plane consists of points $\vec r$ such that the vector $\vec r - \vec a$ is perpendicular to a given normal direction $\vec n$: $(\vec r - \vec a)\cdot\vec n = 0,$ equivalently $\vec r\cdot\vec n = \vec a\cdot\vec n = d.$ Here $\vec n$ is any non-zero normal vector, $\vec a$ is the position vector of any point on the plane, and $d = \vec a\cdot\vec n$ is a constant.

Cartesian (general) form

If $\vec n = (A, B, C)$ and $\vec r = (x, y, z)$, then $\vec r\cdot\vec n = d$ becomes $Ax + By + Cz = D.$ The vector $(A, B, C)$ is normal to the plane. Two planes are parallel iff their normal vectors are proportional; perpendicular iff normals are perpendicular.

Three-point form

A plane through three non-collinear points $A, B, C$ with position vectors $\vec a, \vec b, \vec c$. Any point $\vec r$ on the plane satisfies $\vec{AR}, \vec{AB}, \vec{AC}$ are coplanar: $[\vec r - \vec a\;\;\vec b - \vec a\;\;\vec c - \vec a] = 0.$

In Cartesian form, writing out the triple-product determinant: $\begin{vmatrix} x - x_1 & y - y_1 & z - z_1 \\ x_2 - x_1 & y_2 - y_1 & z_2 - z_1 \\ x_3 - x_1 & y_3 - y_1 & z_3 - z_1 \end{vmatrix} = 0.$

The normal to such a plane is $(\vec b - \vec a)\times(\vec c - \vec a)$.

Intercept form

If a plane has $x$-, $y$-, $z$-intercepts $a, b, c$ (i.e. it crosses the axes at $(a, 0, 0)$, $(0, b, 0)$, $(0, 0, c)$), then it has equation $\frac{x}{a} + \frac{y}{b} + \frac{z}{c} = 1.$ Useful for problems where intercepts are given.

Normal form (Hesse form)

If $\hat n$ is a unit normal to the plane and $p \ge 0$ is the perpendicular distance from the origin to the plane, then $\vec r \cdot \hat n = p.$ The triple $\hat n = (l, m, n)$ contains the direction cosines of the perpendicular from the origin.

To convert $Ax + By + Cz = D$ to normal form: divide by $\sqrt{A^2 + B^2 + C^2}$ (with sign chosen so $D / \sqrt{\cdots} \ge 0$). Then $p = |D|/\sqrt{A^2 + B^2 + C^2}$ and direction cosines are $A/\sqrt{\cdots}$, etc.

Worked examples

Example 1. Find the equation of the plane through $(1, 2, -1)$ with normal $\hat n = (2, 1, 1)$.

$(\vec r - \vec a)\cdot \vec n = 0 \Rightarrow 2(x - 1) + (y - 2) + (z + 1) = 0 \Rightarrow 2x + y + z = 3$.

Example 2. Find the plane through $(1, 0, 0)$, $(0, 1, 0)$, $(0, 0, 1)$.

By intercept form (intercepts 1, 1, 1): $x + y + z = 1$. (Or by three-point form: same answer.)

Example 3. Find the plane through $(2, 1, 0)$, $(3, -2, -2)$, $(3, 1, 7)$.

$\vec{AB} = (1, -3, -2)$, $\vec{AC} = (1, 0, 7)$. Normal $= \vec{AB}\times\vec{AC} = \begin{vmatrix}\hat i & \hat j & \hat k\\1 & -3 & -2\\1 & 0 & 7\end{vmatrix} = \hat i(-21 - 0) - \hat j(7 + 2) + \hat k(0 + 3) = -21\hat i - 9\hat j + 3\hat k$. Simplify to $(-7, -3, 1)$ or $(7, 3, -1)$. Through $(2, 1, 0)$: $7(x - 2) + 3(y - 1) - (z - 0) = 0$, i.e. $7x + 3y - z = 17$.

Example 4. Convert $2x - y + 2z + 6 = 0$ to normal (Hesse) form.

Normal $(2, -1, 2)$, magnitude $\sqrt{4 + 1 + 4} = 3$. Rewrite $2x - y + 2z = -6$, divide by $-3$ (sign flip to make RHS positive): $-\dfrac{2}{3}x + \dfrac{1}{3}y - \dfrac{2}{3}z = 2$. Direction cosines $(-2/3, 1/3, -2/3)$, $p = 2$.

Example 5. Find the plane parallel to $x + 2y - 3z = 5$ and passing through $(1, -1, 2)$.

Same normal $(1, 2, -3)$. Plug in: $1 - 2 - 6 = -7$. So $x + 2y - 3z = -7$.

Example 6. Find the equation of a plane whose intercepts on the axes are $3, -2, 4$.

$\dfrac{x}{3} + \dfrac{y}{-2} + \dfrac{z}{4} = 1 \Rightarrow 4x - 6y + 3z = 12$.

Try it yourself

1. Find the plane through $(0, 0, 0)$ with normal $(1, 2, 3)$.
2. Find the plane through $(1, 1, 1)$, $(2, -1, 3)$, $(0, 1, 2)$.
3. Find intercept form for $2x + 3y + z = 6$.
4. Equation of the $xy$-plane?
5. Plane through the $z$-axis containing the point $(2, 3, 4)$.
6. Convert $x + y + z = \sqrt 3$ to normal form.
7. Find $\lambda$ if $x + 2y + 3z = \lambda$ contains the point $(1, 2, 0)$.
8. Plane parallel to $2x - y + z = 0$ at distance $3$ from origin.
9. Plane through $(1, 0, -1)$ and parallel to $\vec r = (1, 2, 3) + t(1, 1, 0)$ and $\vec r = (2, 1, 0) + s(0, 1, 1)$ , i.e. containing two given direction vectors.
10. Find the equation of the plane through $(2, -1, 3)$, $(1, 2, -1)$, and parallel to the line $\dfrac{x - 1}{1} = \dfrac{y}{2} = \dfrac{z + 1}{-1}$.
11. Show that the planes $x + 2y - 3z = 5$ and $-3x - 6y + 9z = 1$ are parallel.
12. Plane that passes through origin and contains the line $\dfrac{x - 1}{2} = \dfrac{y + 1}{3} = \dfrac{z - 2}{-1}$.
13. Convert vector equation $\vec r\cdot(2\hat i - 3\hat j + 4\hat k) = 5$ to Cartesian.
14. Find the equation of the plane that perpendicularly bisects the segment from $(1, 2, 3)$ to $(3, -2, 5)$.

Pitfalls and tricks

• Normal vector determines a plane up to translation. The constant term encodes the plane's distance from origin (along the normal).
• Three-point form uses the triple product to encode coplanarity.
• Convert between forms freely. Cartesian is concise; vector is conceptually clean; intercept is handy when intercepts are given.
• Normal form requires a unit normal and a non-negative distance. Adjust sign accordingly.
• Parallel planes share a normal direction. Two planes are parallel iff their normal vectors are proportional.